8
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Every letter stands for a digit in decimal representation, different letters stand for different digits:

             SUN
     +      LOSE
     +     UNTIE
     +    BOTTLE
     +   ELISION
     +  NINETEEN
     + NONENTITY
     + EBULLIENT
     + INSOLUBLE
  ------------------
      NEBULOSITY

Which digit does each letter represent?
(Please present the full analysis how these digits can be determined.)

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4
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In each column, the result digit other than N occurs in the sum. Taking the common corresponding digits out of both parts doesn't change the equation or the mapping of letters to digits, but it leaves a convenient result: 1000000000. Let's work with this variant.

Number the columns 0 to 9 from left to right. E.g. SUN contains an N in column 9. Let c$i$ be the carry into column $i$. For example, if column 9 summed to 20, we carry 2 into column 8, so c$8$=2.

Working our way from columns 1 to 9 (c0=N from column 0):

[1] N + I + c1 = 10N [2] 2N + O + c2 = 10c1
[3] E + I + N + S + c3 = 10c2
[4] B + L + N + E + O + c4 = 10c3
[5] 2L + U + I + E + N + c5 = 10c4
[6] 3T + L + N + I + U + c6 = 10c5
[7] 2T + 2E + S + O + I + B + c7 = 10c6
[8] 2L + U + S + I + O + E + N + c8 = 10c7
[9] 4E + 3N + T = 10c8

Consider [9]: LHS min, max respectively occur when (E,N,T) = (0,1,2) and (9,8,7), i.e. 5 $\leq$ LHS $\leq$ 67. But RHS is a multiple of 10, making LHS 10, 20, 30, 40, 50 or 60, so c8 $\in$ [1,6]. Working through each equation independently but carrying the c$i$ iteratively from [9] to [2], we have:

     c8 c7 c6 c5 c4 c3 c2 c1 c0=N
min: 1  3  2  2  2  2  1  1  1
max: 6  5  6  5  4  3  3  2  1

S,L,U,B,E,N are all nonzero because they appear at the start of a word, so one of O,I,T,Y is zero.

We have N=1 from the 'carry' table above.
From [1], I + c1 = 9, so I=7 or I=8, i.e. I is nonzero.
From [9], T must be odd and hence nonzero.
From [2], O must be at least 5 and hence nonzero.
So by elimination, Y=0.

Revising c$i$, we get:

     c8 c7 c6 c5 c4 c3 c2 c1 c0=N
min: 2  4  4  3  2  2  2  1  1
max: 4  5  6  5  4  3  2  1  1

Note that c1 and c2 are now determined, so from [1] I=8, and from [2] O=6.

Set E to 2,3,4,5,7,9 in turn to find simultaneously valid but different S in [3], T in [9] and B in [7]. (E,S,T,B) can only be (2,7,9,3), (5,3,7,4) or (5,4,7,3).

Substituting each (E,S,T,B) in turn into [4] to get L and [5] to get U, we get (E,S,T,B,L,U) = (2,7,9,3,5,4) or (5,3,7,4,2,9). (The case with B=4 is rejected because c4=2 in [4] and c4=3 in [5].)

Substituting each into [6], the case with U=9 requires c6=9 (reject - see 'carry' table). The other case has c6=5, which is legal.

We therefore end up with NEBULOSITY = 1234567890.

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5
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After a bunch of scribbling and start-all-overs:

Y = 0, N = 1, E = 2, B = 3, U = 4, L = 5, O = 6, S = 7, I = 8, T = 9

              741
      +      5672
      +     41982
      +    369952
      +   2587861
      +  18129221
      + 161219890
      + 234558219
      + 817654352
   ------------------
       1234567890

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